0.001 111 122 5 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.001 111 122 5(10) to 64 bit double precision IEEE 754 binary floating point representation standard (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

What are the steps to convert decimal number
0.001 111 122 5(10) to 64 bit double precision IEEE 754 binary floating point representation (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. First, convert to binary (in base 2) the integer part: 0.
Divide the number repeatedly by 2.

Keep track of each remainder.

We stop when we get a quotient that is equal to zero.


  • division = quotient + remainder;
  • 0 ÷ 2 = 0 + 0;

2. Construct the base 2 representation of the integer part of the number.

Take all the remainders starting from the bottom of the list constructed above.

0(10) =

0(2)


3. Convert to binary (base 2) the fractional part: 0.001 111 122 5.

Multiply it repeatedly by 2.

Keep track of each integer part of the results.

Stop when we get a fractional part that is equal to zero.


  • #) multiplying = integer + fractional part;
  • 1) 0.001 111 122 5 × 2 = 0 + 0.002 222 245;
  • 2) 0.002 222 245 × 2 = 0 + 0.004 444 49;
  • 3) 0.004 444 49 × 2 = 0 + 0.008 888 98;
  • 4) 0.008 888 98 × 2 = 0 + 0.017 777 96;
  • 5) 0.017 777 96 × 2 = 0 + 0.035 555 92;
  • 6) 0.035 555 92 × 2 = 0 + 0.071 111 84;
  • 7) 0.071 111 84 × 2 = 0 + 0.142 223 68;
  • 8) 0.142 223 68 × 2 = 0 + 0.284 447 36;
  • 9) 0.284 447 36 × 2 = 0 + 0.568 894 72;
  • 10) 0.568 894 72 × 2 = 1 + 0.137 789 44;
  • 11) 0.137 789 44 × 2 = 0 + 0.275 578 88;
  • 12) 0.275 578 88 × 2 = 0 + 0.551 157 76;
  • 13) 0.551 157 76 × 2 = 1 + 0.102 315 52;
  • 14) 0.102 315 52 × 2 = 0 + 0.204 631 04;
  • 15) 0.204 631 04 × 2 = 0 + 0.409 262 08;
  • 16) 0.409 262 08 × 2 = 0 + 0.818 524 16;
  • 17) 0.818 524 16 × 2 = 1 + 0.637 048 32;
  • 18) 0.637 048 32 × 2 = 1 + 0.274 096 64;
  • 19) 0.274 096 64 × 2 = 0 + 0.548 193 28;
  • 20) 0.548 193 28 × 2 = 1 + 0.096 386 56;
  • 21) 0.096 386 56 × 2 = 0 + 0.192 773 12;
  • 22) 0.192 773 12 × 2 = 0 + 0.385 546 24;
  • 23) 0.385 546 24 × 2 = 0 + 0.771 092 48;
  • 24) 0.771 092 48 × 2 = 1 + 0.542 184 96;
  • 25) 0.542 184 96 × 2 = 1 + 0.084 369 92;
  • 26) 0.084 369 92 × 2 = 0 + 0.168 739 84;
  • 27) 0.168 739 84 × 2 = 0 + 0.337 479 68;
  • 28) 0.337 479 68 × 2 = 0 + 0.674 959 36;
  • 29) 0.674 959 36 × 2 = 1 + 0.349 918 72;
  • 30) 0.349 918 72 × 2 = 0 + 0.699 837 44;
  • 31) 0.699 837 44 × 2 = 1 + 0.399 674 88;
  • 32) 0.399 674 88 × 2 = 0 + 0.799 349 76;
  • 33) 0.799 349 76 × 2 = 1 + 0.598 699 52;
  • 34) 0.598 699 52 × 2 = 1 + 0.197 399 04;
  • 35) 0.197 399 04 × 2 = 0 + 0.394 798 08;
  • 36) 0.394 798 08 × 2 = 0 + 0.789 596 16;
  • 37) 0.789 596 16 × 2 = 1 + 0.579 192 32;
  • 38) 0.579 192 32 × 2 = 1 + 0.158 384 64;
  • 39) 0.158 384 64 × 2 = 0 + 0.316 769 28;
  • 40) 0.316 769 28 × 2 = 0 + 0.633 538 56;
  • 41) 0.633 538 56 × 2 = 1 + 0.267 077 12;
  • 42) 0.267 077 12 × 2 = 0 + 0.534 154 24;
  • 43) 0.534 154 24 × 2 = 1 + 0.068 308 48;
  • 44) 0.068 308 48 × 2 = 0 + 0.136 616 96;
  • 45) 0.136 616 96 × 2 = 0 + 0.273 233 92;
  • 46) 0.273 233 92 × 2 = 0 + 0.546 467 84;
  • 47) 0.546 467 84 × 2 = 1 + 0.092 935 68;
  • 48) 0.092 935 68 × 2 = 0 + 0.185 871 36;
  • 49) 0.185 871 36 × 2 = 0 + 0.371 742 72;
  • 50) 0.371 742 72 × 2 = 0 + 0.743 485 44;
  • 51) 0.743 485 44 × 2 = 1 + 0.486 970 88;
  • 52) 0.486 970 88 × 2 = 0 + 0.973 941 76;
  • 53) 0.973 941 76 × 2 = 1 + 0.947 883 52;
  • 54) 0.947 883 52 × 2 = 1 + 0.895 767 04;
  • 55) 0.895 767 04 × 2 = 1 + 0.791 534 08;
  • 56) 0.791 534 08 × 2 = 1 + 0.583 068 16;
  • 57) 0.583 068 16 × 2 = 1 + 0.166 136 32;
  • 58) 0.166 136 32 × 2 = 0 + 0.332 272 64;
  • 59) 0.332 272 64 × 2 = 0 + 0.664 545 28;
  • 60) 0.664 545 28 × 2 = 1 + 0.329 090 56;
  • 61) 0.329 090 56 × 2 = 0 + 0.658 181 12;
  • 62) 0.658 181 12 × 2 = 1 + 0.316 362 24;

We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit) and at least one integer that was different from zero => FULL STOP (Losing precision - the converted number we get in the end will be just a very good approximation of the initial one).


4. Construct the base 2 representation of the fractional part of the number.

Take all the integer parts of the multiplying operations, starting from the top of the constructed list above:


0.001 111 122 5(10) =

0.0000 0000 0100 1000 1101 0001 1000 1010 1100 1100 1010 0010 0010 1111 1001 01(2)

5. Positive number before normalization:

0.001 111 122 5(10) =

0.0000 0000 0100 1000 1101 0001 1000 1010 1100 1100 1010 0010 0010 1111 1001 01(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 10 positions to the right, so that only one non zero digit remains to the left of it:


0.001 111 122 5(10) =

0.0000 0000 0100 1000 1101 0001 1000 1010 1100 1100 1010 0010 0010 1111 1001 01(2) =

0.0000 0000 0100 1000 1101 0001 1000 1010 1100 1100 1010 0010 0010 1111 1001 01(2) × 20 =

1.0010 0011 0100 0110 0010 1011 0011 0010 1000 1000 1011 1110 0101(2) × 2-10


7. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

Sign 0 (a positive number)

Exponent (unadjusted): -10

Mantissa (not normalized):
1.0010 0011 0100 0110 0010 1011 0011 0010 1000 1000 1011 1110 0101


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-10 + 2(11-1) - 1 =

(-10 + 1 023)(10) =

1 013(10)


9. Convert the adjusted exponent from the decimal (base 10) to 11 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 1 013 ÷ 2 = 506 + 1;
  • 506 ÷ 2 = 253 + 0;
  • 253 ÷ 2 = 126 + 1;
  • 126 ÷ 2 = 63 + 0;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

1013(10) =

011 1111 0101(2)


11. Normalize the mantissa.

a) Remove the leading (the leftmost) bit, since it's allways 1, and the decimal point, if the case.

b) Adjust its length to 52 bits, only if necessary (not the case here).


Mantissa (normalized) =

1. 0010 0011 0100 0110 0010 1011 0011 0010 1000 1000 1011 1110 0101 =

0010 0011 0100 0110 0010 1011 0011 0010 1000 1000 1011 1110 0101


12. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (11 bits) =
011 1111 0101

Mantissa (52 bits) =
0010 0011 0100 0110 0010 1011 0011 0010 1000 1000 1011 1110 0101


Decimal number 0.001 111 122 5 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 0101 - 0010 0011 0100 0110 0010 1011 0011 0010 1000 1000 1011 1110 0101

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How to convert numbers from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 64 bit double precision IEEE 754 binary floating point:

  • 1. If the number to be converted is negative, start with its the positive version.
  • 2. First convert the integer part. Divide repeatedly by 2 the positive representation of the integer number that is to be converted to binary, until we get a quotient that is equal to zero, keeping track of each remainder.
  • 3. Construct the base 2 representation of the positive integer part of the number, by taking all the remainders from the previous operations, starting from the bottom of the list constructed above. Thus, the last remainder of the divisions becomes the first symbol (the leftmost) of the base two number, while the first remainder becomes the last symbol (the rightmost).
  • 4. Then convert the fractional part. Multiply the number repeatedly by 2, until we get a fractional part that is equal to zero, keeping track of each integer part of the results.
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the multiplying operations, starting from the top of the list constructed above (they should appear in the binary representation, from left to right, in the order they have been calculated).
  • 6. Normalize the binary representation of the number, shifting the decimal mark (the decimal point) "n" positions either to the left, or to the right, so that only one non zero digit remains to the left of the decimal mark.
  • 7. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal mark, if the case) and adjust its length to 52 bits, either by removing the excess bits from the right (losing precision...) or by adding extra bits set on '0' to the right.
  • 9. Sign (it takes 1 bit) is either 1 for a negative or 0 for a positive number.

Example: convert the negative number -31.640 215 from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-31.640 215| = 31.640 215

  • 2. First convert the integer part, 31. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 31 ÷ 2 = 15 + 1;
    • 15 ÷ 2 = 7 + 1;
    • 7 ÷ 2 = 3 + 1;
    • 3 ÷ 2 = 1 + 1;
    • 1 ÷ 2 = 0 + 1;
    • We have encountered a quotient that is ZERO => FULL STOP
  • 3. Construct the base 2 representation of the integer part of the number by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above:

    31(10) = 1 1111(2)

  • 4. Then, convert the fractional part, 0.640 215. Multiply repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:
    • #) multiplying = integer + fractional part;
    • 1) 0.640 215 × 2 = 1 + 0.280 43;
    • 2) 0.280 43 × 2 = 0 + 0.560 86;
    • 3) 0.560 86 × 2 = 1 + 0.121 72;
    • 4) 0.121 72 × 2 = 0 + 0.243 44;
    • 5) 0.243 44 × 2 = 0 + 0.486 88;
    • 6) 0.486 88 × 2 = 0 + 0.973 76;
    • 7) 0.973 76 × 2 = 1 + 0.947 52;
    • 8) 0.947 52 × 2 = 1 + 0.895 04;
    • 9) 0.895 04 × 2 = 1 + 0.790 08;
    • 10) 0.790 08 × 2 = 1 + 0.580 16;
    • 11) 0.580 16 × 2 = 1 + 0.160 32;
    • 12) 0.160 32 × 2 = 0 + 0.320 64;
    • 13) 0.320 64 × 2 = 0 + 0.641 28;
    • 14) 0.641 28 × 2 = 1 + 0.282 56;
    • 15) 0.282 56 × 2 = 0 + 0.565 12;
    • 16) 0.565 12 × 2 = 1 + 0.130 24;
    • 17) 0.130 24 × 2 = 0 + 0.260 48;
    • 18) 0.260 48 × 2 = 0 + 0.520 96;
    • 19) 0.520 96 × 2 = 1 + 0.041 92;
    • 20) 0.041 92 × 2 = 0 + 0.083 84;
    • 21) 0.083 84 × 2 = 0 + 0.167 68;
    • 22) 0.167 68 × 2 = 0 + 0.335 36;
    • 23) 0.335 36 × 2 = 0 + 0.670 72;
    • 24) 0.670 72 × 2 = 1 + 0.341 44;
    • 25) 0.341 44 × 2 = 0 + 0.682 88;
    • 26) 0.682 88 × 2 = 1 + 0.365 76;
    • 27) 0.365 76 × 2 = 0 + 0.731 52;
    • 28) 0.731 52 × 2 = 1 + 0.463 04;
    • 29) 0.463 04 × 2 = 0 + 0.926 08;
    • 30) 0.926 08 × 2 = 1 + 0.852 16;
    • 31) 0.852 16 × 2 = 1 + 0.704 32;
    • 32) 0.704 32 × 2 = 1 + 0.408 64;
    • 33) 0.408 64 × 2 = 0 + 0.817 28;
    • 34) 0.817 28 × 2 = 1 + 0.634 56;
    • 35) 0.634 56 × 2 = 1 + 0.269 12;
    • 36) 0.269 12 × 2 = 0 + 0.538 24;
    • 37) 0.538 24 × 2 = 1 + 0.076 48;
    • 38) 0.076 48 × 2 = 0 + 0.152 96;
    • 39) 0.152 96 × 2 = 0 + 0.305 92;
    • 40) 0.305 92 × 2 = 0 + 0.611 84;
    • 41) 0.611 84 × 2 = 1 + 0.223 68;
    • 42) 0.223 68 × 2 = 0 + 0.447 36;
    • 43) 0.447 36 × 2 = 0 + 0.894 72;
    • 44) 0.894 72 × 2 = 1 + 0.789 44;
    • 45) 0.789 44 × 2 = 1 + 0.578 88;
    • 46) 0.578 88 × 2 = 1 + 0.157 76;
    • 47) 0.157 76 × 2 = 0 + 0.315 52;
    • 48) 0.315 52 × 2 = 0 + 0.631 04;
    • 49) 0.631 04 × 2 = 1 + 0.262 08;
    • 50) 0.262 08 × 2 = 0 + 0.524 16;
    • 51) 0.524 16 × 2 = 1 + 0.048 32;
    • 52) 0.048 32 × 2 = 0 + 0.096 64;
    • 53) 0.096 64 × 2 = 0 + 0.193 28;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 52) and at least one integer part that was different from zero => FULL STOP (losing precision...).
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above:

    0.640 215(10) = 0.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 6. Summarizing - the positive number before normalization:

    31.640 215(10) = 1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 7. Normalize the binary representation of the number, shifting the decimal mark 4 positions to the left so that only one non-zero digit stays to the left of the decimal mark:

    31.640 215(10) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 20 =
    1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

  • 9. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as shown above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1 = (4 + 1023)(10) = 1027(10) =
    100 0000 0011(2)

  • 10. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign) and adjust its length to 52 bits, by removing the excess bits, from the right (losing precision...):

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

    Mantissa (normalized): 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 100 0000 0011

    Mantissa (52 bits) = 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Number -31.640 215, converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point =
    1 - 100 0000 0011 - 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100